MPI_Allreduce 双打

Question

我正在尝试为我正在处理的排序查找所有处理器的全局最小值和最大值。我正在尝试使用 MPI_Reduceall

int rank, nproc;
MPI_Comm_size(MPI_COMM_WORLD,&nproc);
MPI_Comm_rank(MPI_COMM_WORLD,&rank);

vector< vector<double> > buckets(nproc);
double local_min = *std::min_element(values_to_sort.begin(), values_to_sort.end());
double local_max = *std::max_element(values_to_sort.begin(), values_to_sort.end());

int min = 0;
int max = 0;

double global_min;
double global_max;

MPI_Allreduce(&local_min, &global_min, 1, MPI_2DOUBLE_PRECISION, MPI_MINLOC, MPI_COMM_WORLD);
MPI_Allreduce(&local_max, &global_max, 1, MPI_2DOUBLE_PRECISION, MPI_MAXLOC, MPI_COMM_WORLD);

cout << "local_min " << local_min << " local_max " << local_max << endl;
cout << "global_min " << global_min << " global_max " << global_max << endl;

显示了我的代码，但每次都出现段错误。这是一个简单的函数，使用以下方法生成的随机双打调用：

int min = 0;
    int max = 100;

    vector<double> values_to_sort;
    vector<double> sorted_values;
    for(int i=0; i< 1000; i++)
    {
      values_to_sort.push_back( ((double) rand()*(max-min)/(double)RAND_MAX-min) );
    }

如果有人知道这个段错误的原因，请告诉我。我真的很想以一种快速简单的方式获得全局最大值和最小值。

Answer 1

我读了一些书，MINLOC 和 MAXLOC 操作要求您使用某个大小大于 1 的数组来存储值并成对排序。

我更新的代码

 int rank, nproc;
    MPI_Comm_size(MPI_COMM_WORLD,&nproc);
    MPI_Comm_rank(MPI_COMM_WORLD,&rank);

    vector< vector<double> > buckets(nproc);
    double local_min[2];
    local_min[1] = rank;
    local_min[0] = *std::min_element(values_to_sort.begin(), values_to_sort.end());

    double local_max[2];
    local_max[1] = rank;
    local_max[0] = *std::max_element(values_to_sort.begin(), values_to_sort.end());

    double global_min[2];
    double global_max[2];

    MPI_Allreduce(&local_min, &global_min, 1, MPI_2DOUBLE_PRECISION, MPI_MINLOC, MPI_COMM_WORLD);
    MPI_Allreduce(&local_max, &global_max, 1, MPI_2DOUBLE_PRECISION, MPI_MAXLOC, MPI_COMM_WORLD);

    cout << "local_min " << local_min[0] << " local_max " << local_max[0] << endl;
    cout << "global_min " << global_min[0] << " global_max " << global_max[0] << endl;

Answer 2

你真的需要在这里使用 MPI_MINLOC 和 MPI_MAXLOC 吗？因为您陈述问题的方式和使用解决方案的方式，MPI_MIN 和 MPI_MAX 就可以了：

double local_min = *std::min_element(values_to_sort.begin(), values_to_sort.end());
double local_max = *std::max_element(values_to_sort.begin(), values_to_sort.end());

double global_min;
double global_max;

MPI_Allreduce(&local_min, &global_min, 1, MPI_DOUBLE, MPI_MIN, MPI_COMM_WORLD);
MPI_Allreduce(&local_max, &global_max, 1, MPI_DOUBLE, MPI_MAX, MPI_COMM_WORLD);

cout << "local_min " << local_min << " local_max " << local_max << endl;
cout << "global_min " << global_min << " global_max " << global_max << endl;

现在，如果你真的需要拥有全局最小值和全局最大值的进程的等级，那么你确实需要MPI_MINLOC和MPI_MAXLOC。但是，您应该这样使用它：

struct double_int {
    double val;
    int rank;
} local_min, local_max, global_min, global_max;

local_min.val = *std::min_element(values_to_sort.begin(), values_to_sort.end());
local_max.val = *std::max_element(values_to_sort.begin(), values_to_sort.end());
local_min.rank = local_max.rank = rank;

MPI_Allreduce(&local_min, &global_min, 1, MPI_DOUBLE_INT, MPI_MINLOC, MPI_COMM_WORLD);
MPI_Allreduce(&local_max, &global_max, 1, MPI_DOUBLE_INT, MPI_MAXLOC, MPI_COMM_WORLD);

cout << "on process " << rank << " local_min " << local_min.val
     << " local_max " << local_max.val << endl;
cout << "global_min " << global_min.val << " owned by process " << global_min.rank << endl;
cout << "global_max " << global_max.val << " owned by process " << global_max.rank << endl;

MPI_Allreduce 双打

MPI_Allreduce doubles

c++

mpi

openmpi